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2 edited body

The easiest way I know to say what is going on is to resort to looking at "products" of pairs: $$(X, A) \times (Y, B) = ( X\times Y , A\times Y \cup X\times B).$$ The point of this notation is that the functor $(X, A) \mapsto (X/A, *)$ carries $(X, A) \times (Y, B)$ to $X/A \wedge Y/B$. We can iterate this procedure, and I'll write $T^n(Y,X)$ for the subspace of $Y^n$ satisfying $$(Y, X)^n = ( Y^n, T^n(Y, X)).$$ Thus $(Y/X)^{\wedge n} = Y^n /T^n(Y,X)$.

You can easily check that $$T^n( Y, X) = \lbrace (y_1, \ldots, y_n) \mid y_i \in X\ \mbox{for at least one i}\rbrace.$$

On the other hand $Y^{\wedge n}/X^{\wedge n}$ is the quotient of $Y^n$ by the subspace $$T^n(X,*T^n(Y,*) \cup X^n,$$ which is different (unless $X = *$).

1

The easiest way I know to say what is going on is to resort to looking at "products" of pairs: $$(X, A) \times (Y, B) = ( X\times Y , A\times Y \cup X\times B).$$ The point of this notation is that the functor $(X, A) \mapsto (X/A, *)$ carries $(X, A) \times (Y, B)$ to $X/A \wedge Y/B$.
We can iterate this procedure, and I'll write $T^n(Y,X)$ for the subspace of $Y^n$ satisfying $$(Y, X)^n = ( Y^n, T^n(Y, X)).$$ Thus $(Y/X)^{\wedge n} = Y^n /T^n(Y,X)$.

You can easily check that $$T^n( Y, X) = \lbrace (y_1, \ldots, y_n) \mid y_i \in X\ \mbox{for at least one i}\rbrace.$$

On the other hand $Y^{\wedge n}/X^{\wedge n}$ is the quotient of $Y^n$ by the subspace $$T^n(X,*) \cup X^n,$$ which is different (unless $X = *$).